Mathematics Class 11
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Slope of a Line and Angle Between Two Lines

🎓 Class 11 Mathematics CBSE Theory Ch 9 — Straight Lines ⏱ ~30 min
🌐 Language: [gtranslate]

This MCQ module is based on: Slope of a Line and Angle Between Two Lines

This mathematics assessment will be based on: Slope of a Line and Angle Between Two Lines
Targeting Class 11 level in Coordinate Geometry, with Advanced difficulty.

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9.1 Introduction

We are familiar with plotting points in the Cartesian plane and finding the distance between two points. In this chapter we extend these ideas to study the simplest curve in the plane — the straight linei. Every straight line on a coordinate plane can be represented by a first-degree equation in \(x\) and \(y\). We will learn how to characterize a line by its slope, the angle it makes with another line, the different algebraic forms of its equation, and how to measure the perpendicular distance of a point from a line or between two parallel lines.

The analytic (coordinate) treatment of lines owes its origin to the French philosopher–mathematician René Descartes, who showed in 1637 that every algebraic equation in two variables corresponds to a curve in the plane.

René Descartes
(1596 – 1650)
French philosopher and mathematician. His La Géométrie (1637) founded analytic geometry and made possible the algebraic study of straight lines and curves.

9.2 Slope of a Line

A line in the plane makes two angles with the positive direction of the \(x\)-axis which are supplementary to each other. The smaller non-negative angle \(\theta\) (\(0° \le \theta < 180°\)) that the line makes with the positive \(x\)-axis in the anti-clockwise direction is called the inclinationi of the line.

Definition — Slope
If \(\theta\) is the inclination of a line \(l\), then \(\tan\theta\) is called the slope or gradient of \(l\), denoted \(m\). Thus \(m=\tan\theta,\ \theta\neq 90°\). A vertical line has undefined slope; a horizontal line has slope 0.
xy θ P Q line l, slope m = tan θ
Fig 9.1 — The inclination \(\theta\) of line \(l\) with the positive \(x\)-axis; its slope is \(m=\tan\theta\).

9.2.1 Slope of a line through two given points

Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) be two distinct points on a non-vertical line \(l\) with inclination \(\theta\). Drop perpendiculars from \(P\) and \(Q\) to the \(x\)-axis and let them meet at \(M(x_1,0)\) and \(N(x_2,0)\). Drawing \(PR\parallel x\)-axis with \(R\) on \(QN\), we obtain a right triangle \(PRQ\) with:

\(\displaystyle\tan\theta=\frac{RQ}{PR}=\frac{y_2-y_1}{x_2-x_1}\).

Key formula
The slope of the line through \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) (with \(x_1\neq x_2\)) is \[m=\frac{y_2-y_1}{x_2-x_1}.\] The formula is symmetric — swapping the two points gives the same slope.
xy P(x₁,y₁) Q(x₂,y₂) R x₂ − x₁ y₂ − y₁ θ
Fig 9.2 — Slope as rise over run: \(m=\dfrac{y_2-y_1}{x_2-x_1}\).

9.2.2 Conditions for parallelism and perpendicularity

Two non-vertical lines \(l_1\) and \(l_2\) with slopes \(m_1\) and \(m_2\) and inclinations \(\alpha\) and \(\beta\) satisfy:

  • Parallel \(\iff\) \(\alpha=\beta \iff m_1=m_2\).
  • Perpendicular \(\iff\) \(\beta=\alpha+90° \iff \tan\beta=-\cot\alpha\iff m_1 m_2=-1\).

9.2.3 Angle between two lines

Let \(\theta\) be the acute angle between lines \(l_1\) and \(l_2\) with slopes \(m_1\) and \(m_2\) (and \(1+m_1 m_2\neq 0\)). Using \(\tan(\alpha-\beta)\):

\[\tan\theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|.\]

If \(1+m_1m_2=0\), the lines are perpendicular.

l₁ (slope m₁) l₂ (slope m₂) θ
Fig 9.3 — The acute angle \(\theta\) between lines \(l_1\) and \(l_2\).

9.2.4 Collinearity of three points

Three points \(A,B,C\) are collineari if and only if the slope of \(AB\) equals the slope of \(BC\):

\[\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}.\]

Example 1 — Slope computation
Find the slope of the line through \(A(3,-2)\) and \(B(-1,4)\).
\(m=\dfrac{4-(-2)}{-1-3}=\dfrac{6}{-4}=-\dfrac{3}{2}.\)
Example 2 — Parallel lines
Show that the line through \((3,-2)\) and \((-2,6)\) is parallel to the line through \((0,8)\) and \((-5,16)\).
First slope \(=\dfrac{6-(-2)}{-2-3}=-\dfrac{8}{5}\). Second \(=\dfrac{16-8}{-5-0}=-\dfrac{8}{5}\). Equal \(\Rightarrow\) parallel.
Example 3 — Perpendicular lines
If the line through \((4,-2)\) and \((k,-5)\) is perpendicular to the line through \((-1,3)\) and \((1,7)\), find \(k\).
Second slope \(=\dfrac{7-3}{1-(-1)}=2\). First slope \(=\dfrac{-5+2}{k-4}=\dfrac{-3}{k-4}\). Perpendicular \(\Rightarrow\) product \(=-1\): \(\dfrac{-3}{k-4}\cdot 2=-1\Rightarrow k-4=6\Rightarrow k=10.\)
Example 4 — Angle between two lines
Find the acute angle between the lines with slopes \(m_1=\tfrac{1}{2}\) and \(m_2=3\).
\(\tan\theta=\left|\dfrac{3-\tfrac12}{1+\tfrac12\cdot 3}\right|=\left|\dfrac{5/2}{5/2}\right|=1\Rightarrow \theta=45°.\)
Example 5 — Collinearity
Show that \(A(-2,-1), B(4,0), C(3,3)\) are not collinear.
Slope \(AB=\dfrac{0+1}{4+2}=\tfrac{1}{6}\); slope \(BC=\dfrac{3-0}{3-4}=-3\). Since \(\tfrac{1}{6}\neq -3\), the points are not collinear.
Activity 9.1 — Slope detective on a grid
Predict: On a grid, draw a line joining \((1,1)\) and \((4,7)\). Without computing, do you think its slope is closer to 1, 2 or 3?
  1. Compute the slope using the two-point formula.
  2. Mark a third point \((2,3)\). Is it on the line? Check via slopes.
  3. Draw the line through \((0,0)\) perpendicular to this one and state its slope.
  4. Write down the inclination of the original line in degrees (use a calculator).
Insight: Slope \(=\frac{7-1}{4-1}=2\). Checking \((2,3)\): slope from \((1,1)\) is \(\frac{3-1}{2-1}=2\) — yes, collinear. Perpendicular slope \(=-\tfrac12\). Inclination \(=\tan^{-1}2\approx 63.43°\).

9.3 In-text Exercises

Q1. Find the slope of the line through \((3,-4)\) and \((-5,6)\).
\(m=\dfrac{6-(-4)}{-5-3}=\dfrac{10}{-8}=-\dfrac{5}{4}.\)
Q2. Without using Pythagoras, show that \(A(4,4), B(3,5), C(-1,-1)\) are vertices of a right-angled triangle.
Slope \(AB=-1\); slope \(AC=\dfrac{-1-4}{-1-4}=1\); slope \(BC=\dfrac{-1-5}{-1-3}=\dfrac{3}{2}\). Product of slopes \(AB\) and \(AC\) \(=-1\cdot1=-1\), so \(AB\perp AC\). Right angle at \(A\).
Q3. Find \(y\) so that the line through \((2,-3)\) and \((4,y)\) has slope \(2\).
\(2=\dfrac{y+3}{4-2}\Rightarrow y+3=4\Rightarrow y=1.\)
Q4. Find the angle between the lines with slopes \(m_1=1\) and \(m_2=-1\).
Since \(1+m_1m_2=1-1=0\), the lines are perpendicular; \(\theta=90°\).
Q5. Check whether \(P(-2,3), Q(1,2), R(4,1)\) are collinear.
Slope \(PQ=\dfrac{2-3}{1+2}=-\tfrac13\); slope \(QR=\dfrac{1-2}{4-1}=-\tfrac13\). Equal \(\Rightarrow\) collinear.
Competency-Based Questions — Slope in the Real World
A highway designer plots the route of a straight ramp on an engineer's Cartesian map. The ramp starts at \(A(2,1)\) (in kilometres) and ends at \(B(10,5)\). A second proposed service road runs through \(C(4,8)\) and \(D(8,6)\).
Q1. Compute the slope of the main ramp \(AB\) and interpret its physical meaning.
L3 Apply
\(m_{AB}=\dfrac{5-1}{10-2}=\dfrac{1}{2}\). For every 2 km east, the ramp rises 1 km north on the map — a gentle gradient.
Q2. Determine whether the ramp \(AB\) is perpendicular to the service road \(CD\). Show working.
L4 Analyse
\(m_{CD}=\dfrac{6-8}{8-4}=-\dfrac{1}{2}\). Product \(=\tfrac12\cdot(-\tfrac12)=-\tfrac14\neq -1\). Not perpendicular.
Q3. The builder claims that a third stretch joining \(A(2,1)\), \(M(6,3)\), \(B(10,5)\) forms one continuous straight ramp. Evaluate the claim.
L5 Evaluate
Slope \(AM=\dfrac{3-1}{6-2}=\tfrac12\) and slope \(MB=\dfrac{5-3}{10-6}=\tfrac12\). Equal \(\Rightarrow\) \(A,M,B\) collinear. Claim valid.
Q4. Design an additional road through \(C(4,8)\) that is perpendicular to the ramp \(AB\). Give two points on it and justify.
L6 Create
Perpendicular slope \(=-2\). Starting at \(C(4,8)\), a second point can be \((5,6)\) since \(\dfrac{6-8}{5-4}=-2\). Product \(\tfrac12\cdot(-2)=-1\) confirms perpendicularity.
Assertion–Reason Questions
Assertion (A): The slope of a horizontal line is 0.
Reason (R): For a horizontal line, the inclination \(\theta=0°\) and \(\tan 0°=0\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. Slope \(=\tan\theta\); horizontal line has \(\theta=0°\), so \(m=0\).
Assertion (A): If three points \(A,B,C\) satisfy slope\((AB)=\) slope\((BC)\), they must be collinear.
Reason (R): Equal slopes at a common point \(B\) imply lines \(AB\) and \(BC\) coincide with the same line through \(B\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. Two lines through the same point with the same slope coincide; hence \(A,B,C\) lie on one line.
Assertion (A): If the slopes of two lines are \(m_1=3\) and \(m_2=-\tfrac13\), the lines are perpendicular.
Reason (R): Two lines are perpendicular when the sum of their slopes is \(-1\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: C. A is true because \(3\times(-\tfrac13)=-1\). R is false: the product, not the sum, must equal \(-1\).

Frequently Asked Questions

What is the slope formula for two points?
The slope of the line through (x1, y1) and (x2, y2) is m = (y2 - y1) / (x2 - x1), provided x1 is not equal to x2.
What is the slope of a horizontal line?
A horizontal line has slope 0 because its inclination is 0 degrees and tan 0 = 0.
What is the slope of a vertical line?
A vertical line has an undefined slope because its inclination is 90 degrees and tan 90 is not defined.
What is the condition for two lines to be parallel?
Two non-vertical lines are parallel if and only if their slopes are equal: m1 = m2.
What is the condition for two lines to be perpendicular?
Two non-vertical lines are perpendicular if and only if the product of their slopes is -1: m1 x m2 = -1.
What is the angle between two lines formula?
If m1 and m2 are the slopes, the acute angle theta between the lines satisfies tan theta = |(m2 - m1)/(1 + m1 m2)|, provided 1 + m1 m2 is not 0.
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